If the distinct nonzero numbers $x(y - z)$, $y(z - x)$, $z(x - y)$ form a geometric progression with common ratio $r$, then what equation does $r$ satisfy?
I set $r$ = $\frac{y(z-x)}{x(y-z)}$ = $\frac{z(x-y)}{y(z-x)}$ and then cross multiplied and tried to simplify the algebra but I couldn't get it to work out.
We have $$y(z-x)=rx(y-z)\tag1$$ $$z(x-y)=ry(z-x)\tag2$$
From $(1)$, $$z(y+rx)=xy(r+1)\tag3$$ From $(2)$, $$z(x-y-ry)=-xyr\tag4$$
From $(3)(4)$, we have $$(z(y+rx)(x-y-ry)=)\quad xy(r+1)(x-y-ry)=-xyr(y+rx)$$ Dividing the both sides by $xy$ gives $$(r+1)(x-y-ry)=-r(y+rx),$$ i.e. $$(r+1)(x-y)-r(r+1)y+ry+r^2x=0$$ i.e. $$(r+1)(x-y)+r^2(x-y)=0$$ $$(r^2+r+1)(x-y)=0$$ Since $x-y\not=0$, we obtain $$\color{red}{r^2+r+1=0}.$$